Temperature of ice point T_{1} = 273.15 K

Temperature of steam point T_{2} = 372.15 K

Pressure of the gas in a constant volume thermometer at ice point P_{1} = 70 kpa

Now,

Let Ptr be the pressure of the triple point and P_{2} be the pressure at the steam point.

The temperature - pressure relations for ice point and steam point are given as follows:-

For Ice Point,

T_{1 }= P_{1}/Ptr × 273.16 K

273.15 = 70/Ptr × 10×10×10×273.16

Ptr = (70×10×10×10×273.16)/273.15 Pa

For Steam Point,

T_{2} = 273.16/Ptr K

On substituting the value of Ptr, we get:

373.15 = (P_{2} × 273.15 × 273.16)/(70 × 273.16 × 10 × 10 × 10)

P_{2} = 95.626 × 1000 Pa

P_{2} = 96kpa (approx.)

**Therefore, the pressure at steam point is 96kpa**